Question

Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature, are given below. reaction (1): N2(g) + O2(g) 2 NO(g); Kc = 1.54e-31 reaction (2): N2(g) + 1/2 O2(g) N2O(g); Kc = 2.61e-24 Using this set of data, determine the equilibrium constant for the following reaction, at the same temperature. reaction (3): N2O(g) + 1/2 O2(g) 2 NO(g)

Answer 1

Kc = 1.54e - 31 / 2.61e - 24

Step-by-step explanation:

1 )
`N_(2)(gas) + O_(2)(gas)\rightarrow 2NO(gas)` ; Kc = 1.54e - 31

2)
`N_(2)(gas) + 1/2O_(2)(gas)\rightarrow N_(2)O(gas)` ; Kc = 2.16e - 24

upon reversing ( 2 ) equation

`N_(2)O(gas)\rightarrow N_(2)(gas) + 1/2O_(2)(gas)` Kc = 1/2.16e - 24

now adding 1 and reversed equation (2)

`N_(2)(gas) + O_(2)(gas)\rightarrow 2NO(gas)`

`N_(2)O(gas)\rightarrow N_(2)(gas) + 1/2O_(2)(gas)`

we get ,

`N_(2)O(gas) + 1/2O_(2)(gas)\rightarrow 2NO(gas)` Kc = 1.54e-31 × 1/2.61e - 24

equilibrium constant of equation (3) is -

Kc = 1.54e - 31 / 2.61e - 24